Physics Past Year Paper Theory (Outside Delhi Set -1) - 2015, Class 12, CBSE Class 12 Notes | EduRev

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Class 12 : Physics Past Year Paper Theory (Outside Delhi Set -1) - 2015, Class 12, CBSE Class 12 Notes | EduRev

 Page 1


55/1 1 [P.T.O. 
¸üÖê»Ö ®ÖÓ.
Roll No. 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70
Time allowed : 3 hours ] [ Maximum Marks : 70 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :
(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü …  
(ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ … 
(iii) ÜÖÞ›ü-†ú ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛ  Ûêú 5 †ÓÛú Æïü … 
(iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …
 Series : SSO/C 
55/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12 Æïü …
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …
• Please check that this question paper contains 12 printed pages.
• Code number given on the right hand side of the question paper should be written on the
title page of the answer-book by the candidate.
• Please check that this question paper contains 26 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes time has been allotted to read this question paper. The question paper will be
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the
question paper only and will not write any answer on the answer-book during this period.
ÛúÖê›ü ®ÖÓ. 
Code No. 
   
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book.
SET – 1 
Page 2


55/1 1 [P.T.O. 
¸üÖê»Ö ®ÖÓ.
Roll No. 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70
Time allowed : 3 hours ] [ Maximum Marks : 70 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :
(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü …  
(ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ … 
(iii) ÜÖÞ›ü-†ú ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛ  Ûêú 5 †ÓÛú Æïü … 
(iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …
 Series : SSO/C 
55/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12 Æïü …
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …
• Please check that this question paper contains 12 printed pages.
• Code number given on the right hand side of the question paper should be written on the
title page of the answer-book by the candidate.
• Please check that this question paper contains 26 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes time has been allotted to read this question paper. The question paper will be
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the
question paper only and will not write any answer on the answer-book during this period.
ÛúÖê›ü ®ÖÓ. 
Code No. 
   
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book.
SET – 1 
55/1 2 
(v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü :
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg
¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg
†Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö 
²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
General Instructions : 
(i) All questions are compulsory. There are 26 questions in all.  
(ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
(iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
(iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
(v) You may use the following values of physical constants wherever necessary : 
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
mass of neutron = 1.675 × 10
–27
 kg 
mass of proton = 1.673 × 10
–27
 kg 
Avogadro’s number = 6.023 × 10
23
 per gram mole 
Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
Page 3


55/1 1 [P.T.O. 
¸üÖê»Ö ®ÖÓ.
Roll No. 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70
Time allowed : 3 hours ] [ Maximum Marks : 70 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :
(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü …  
(ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ … 
(iii) ÜÖÞ›ü-†ú ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛ  Ûêú 5 †ÓÛú Æïü … 
(iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …
 Series : SSO/C 
55/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12 Æïü …
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …
• Please check that this question paper contains 12 printed pages.
• Code number given on the right hand side of the question paper should be written on the
title page of the answer-book by the candidate.
• Please check that this question paper contains 26 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes time has been allotted to read this question paper. The question paper will be
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the
question paper only and will not write any answer on the answer-book during this period.
ÛúÖê›ü ®ÖÓ. 
Code No. 
   
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book.
SET – 1 
55/1 2 
(v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü :
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg
¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg
†Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö 
²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
General Instructions : 
(i) All questions are compulsory. There are 26 questions in all.  
(ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
(iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
(iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
(v) You may use the following values of physical constants wherever necessary : 
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
mass of neutron = 1.675 × 10
–27
 kg 
mass of proton = 1.673 × 10
–27
 kg 
Avogadro’s number = 6.023 × 10
23
 per gram mole 
Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1 3 [P.T.O. 
ÜÖÞ›ü – †
Section – A 
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1 
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1 
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1 
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1 
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1 
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B 
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö 
?
E = 3 × 10
3
 
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê 
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
 
^
i N/C through a square of side 10 cm, when it is held normal to 
?
E ? 
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the 
figure. 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Page 4


55/1 1 [P.T.O. 
¸üÖê»Ö ®ÖÓ.
Roll No. 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70
Time allowed : 3 hours ] [ Maximum Marks : 70 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :
(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü …  
(ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ … 
(iii) ÜÖÞ›ü-†ú ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛ  Ûêú 5 †ÓÛú Æïü … 
(iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …
 Series : SSO/C 
55/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12 Æïü …
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …
• Please check that this question paper contains 12 printed pages.
• Code number given on the right hand side of the question paper should be written on the
title page of the answer-book by the candidate.
• Please check that this question paper contains 26 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes time has been allotted to read this question paper. The question paper will be
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the
question paper only and will not write any answer on the answer-book during this period.
ÛúÖê›ü ®ÖÓ. 
Code No. 
   
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book.
SET – 1 
55/1 2 
(v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü :
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg
¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg
†Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö 
²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
General Instructions : 
(i) All questions are compulsory. There are 26 questions in all.  
(ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
(iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
(iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
(v) You may use the following values of physical constants wherever necessary : 
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
mass of neutron = 1.675 × 10
–27
 kg 
mass of proton = 1.673 × 10
–27
 kg 
Avogadro’s number = 6.023 × 10
23
 per gram mole 
Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1 3 [P.T.O. 
ÜÖÞ›ü – †
Section – A 
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1 
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1 
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1 
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1 
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1 
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B 
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö 
?
E = 3 × 10
3
 
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê 
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
 
^
i N/C through a square of side 10 cm, when it is held normal to 
?
E ? 
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the 
figure. 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
55/1 4 
8. 20 cm ³Öã•ÖÖ ¾ÖÖ»Öê ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü »Öæ¯Ö וÖÃÖÃÖê 1A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü, ÛúÖê ×ÛúÃÖß †®Ö®ŸÖ »Ö´²ÖÖ‡Ô Ûêú ÃÖ߬Öê
ŸÖÖ¸ü וÖÃÖÃÖê 2A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü Ûêú ×®ÖÛú™ü ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÃÖ´ÖÖ®Ö ŸÖ»Ö ´Öë ¸üÜÖÖ ÝÖµÖÖ Æîü … 2 
¬ÖÖ¸üÖ¾ÖÖÆüß “ÖÖ»ÖÛú Ûêú ÛúÖ¸üÞÖ »Öæ¯Ö ¯Ö¸ü †Ö¸üÖê×¯ÖŸÖ ®Öê™ü ²Ö»Ö ÛúÖ ¯Ö׸ü´ÖÖÞÖ †Öî¸ü פü¿ÖÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
†£Ö¾ÖÖ 
200 ±êú¸üÖë †Öî¸ü 100 cm
2 
õÖê¡Ö±ú»Ö Ûúß ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü ÃÖ´ÖŸÖ»Ö ÛãúÞ›ü»Öß ÃÖê 5A †¯Ö׸ü¾ÖŸÖá ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß 
Æîü … µÖÆü ÛãúÞ›ü»Öß 0.2 T Ûêú ‹êÃÖê ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ´Öë ×Ã£ÖŸÖ Æîü, וÖÃÖÛúß ×¤ü¿ÖÖ ÛãúÞ›ü»Öß Ûêú ŸÖ»Ö Ûêú 
»Ö´²Ö¾ÖŸÖ Æîü … •Ö²Ö ‡ÃÖ ÛãúÞ›ü»Öß ÛúÖ ŸÖ»Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ÃÖê 60° ÛúÖ ÛúÖêÞÖ ²Ö®ÖÖŸÖÖ Æîü ŸÖ²Ö ˆÃÖ ×ãÖ×ŸÖ ´Öë ÛãúÞ›ü»Öß 
¯Ö¸ü »ÖÝÖÖ ²Ö»Ö-†Ö‘ÖæÞÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … ×ÛúÃÖ ×¾Ö®µÖÖÃÖ ´Öë µÖÆü ÛãúÞ›ü»Öß Ã£ÖÖµÖß ÃÖÖ´µÖÖ¾ÖãÖÖ ´Öë ÆüÖêÝÖß ?
A square loop of side 20 cm carrying current of 1A is kept near an infinite long 
straight wire carrying a current of 2A in the same plane as shown in the figure. 
Calculate the magnitude and direction of the net force exerted on the loop due to the 
current carrying conductor. 
OR 
A square shaped plane coil of area 100 cm
2
 of 200 turns carries a steady current of 5A. 
It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of 
the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the 
direction of the field. In which orientation will the coil be in stable equilibrium ? 
9. ˆ®Ö ¾ÖîªãŸÖ “Öã´²ÖÛúßµÖ ×¾Ö×Ûú¸üÞÖÖë ÛúÖ ®ÖÖ´Ö ×»Ö×ÜÖ‹ (i) ו֮ÖÛúÖ ˆ¯ÖµÖÖêÝÖ ÛïúÃÖ¸ü Ûúß ÛúÖê׿ÖÛúÖ†Öë ÛúÖê ®Ö™ü Ûú¸ü®Öê ´Öë
×ÛúµÖÖ •ÖÖŸÖÖ Æîü, (ii) ו֮ÖÃÖê Ÿ¾Ö“ÖÖ ŸÖÖ´ÖÏ ¸ÓüÝÖ Ûúß ÆüÖê •ÖÖŸÖß Æîü, (iii) ¯Ö飾Öß Ûúß ˆÂÞÖŸÖÖ ²Ö®ÖÖ‹ ¸üÜÖŸÖê Æïü … 2
‡®Ö´Öë ÃÖê ×ÛúÃÖß ‹Ûú ¯ÖÏÛúÖ¸ü Ûúß ŸÖ¸ÓüÝÖÖë ÛúÖê ˆŸ¯Ö®®Ö Ûú¸ü®Öê Ûúß ×¾Ö×¬Ö ÛúÖ ÃÖÓõÖê¯Ö ´Öë ¾ÖÞÖÔ®Ö Ûúßו֋ …
Name the types of e.m. radiations which (i) are used in destroying cancer cells,
(ii) cause tanning of the skin and (iii) maintain the earth’s warmth.
Write briefly a method of producing any one of these waves.
Page 5


55/1 1 [P.T.O. 
¸üÖê»Ö ®ÖÓ.
Roll No. 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70
Time allowed : 3 hours ] [ Maximum Marks : 70 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :
(i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü …  
(ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ … 
(iii) ÜÖÞ›ü-†ú ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛ  Ûêú 5 †ÓÛú Æïü … 
(iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …
 Series : SSO/C 
55/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12 Æïü …
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …
• Please check that this question paper contains 12 printed pages.
• Code number given on the right hand side of the question paper should be written on the
title page of the answer-book by the candidate.
• Please check that this question paper contains 26 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes time has been allotted to read this question paper. The question paper will be
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the
question paper only and will not write any answer on the answer-book during this period.
ÛúÖê›ü ®ÖÓ. 
Code No. 
   
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book.
SET – 1 
55/1 2 
(v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü :
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg
¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg
†Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö 
²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
General Instructions : 
(i) All questions are compulsory. There are 26 questions in all.  
(ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
(iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
(iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
(v) You may use the following values of physical constants wherever necessary : 
c = 3 × 10
8
 m/s 
h = 6.63 × 10
–34
 Js 
e = 1.6 × 10
–19
 C 
µ
0
 = 4p × 10
–7
 T m A
–1
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2
m
e
 = 9.1 × 10
–31
 kg
mass of neutron = 1.675 × 10
–27
 kg 
mass of proton = 1.673 × 10
–27
 kg 
Avogadro’s number = 6.023 × 10
23
 per gram mole 
Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1 3 [P.T.O. 
ÜÖÞ›ü – †
Section – A 
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1 
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1 
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1 
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1 
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1 
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B 
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö 
?
E = 3 × 10
3
 
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê 
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
 
^
i N/C through a square of side 10 cm, when it is held normal to 
?
E ? 
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the 
figure. 
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
55/1 4 
8. 20 cm ³Öã•ÖÖ ¾ÖÖ»Öê ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü »Öæ¯Ö וÖÃÖÃÖê 1A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü, ÛúÖê ×ÛúÃÖß †®Ö®ŸÖ »Ö´²ÖÖ‡Ô Ûêú ÃÖ߬Öê
ŸÖÖ¸ü וÖÃÖÃÖê 2A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü Ûêú ×®ÖÛú™ü ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÃÖ´ÖÖ®Ö ŸÖ»Ö ´Öë ¸üÜÖÖ ÝÖµÖÖ Æîü … 2 
¬ÖÖ¸üÖ¾ÖÖÆüß “ÖÖ»ÖÛú Ûêú ÛúÖ¸üÞÖ »Öæ¯Ö ¯Ö¸ü †Ö¸üÖê×¯ÖŸÖ ®Öê™ü ²Ö»Ö ÛúÖ ¯Ö׸ü´ÖÖÞÖ †Öî¸ü פü¿ÖÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
†£Ö¾ÖÖ 
200 ±êú¸üÖë †Öî¸ü 100 cm
2 
õÖê¡Ö±ú»Ö Ûúß ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü ÃÖ´ÖŸÖ»Ö ÛãúÞ›ü»Öß ÃÖê 5A †¯Ö׸ü¾ÖŸÖá ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß 
Æîü … µÖÆü ÛãúÞ›ü»Öß 0.2 T Ûêú ‹êÃÖê ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ´Öë ×Ã£ÖŸÖ Æîü, וÖÃÖÛúß ×¤ü¿ÖÖ ÛãúÞ›ü»Öß Ûêú ŸÖ»Ö Ûêú 
»Ö´²Ö¾ÖŸÖ Æîü … •Ö²Ö ‡ÃÖ ÛãúÞ›ü»Öß ÛúÖ ŸÖ»Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ÃÖê 60° ÛúÖ ÛúÖêÞÖ ²Ö®ÖÖŸÖÖ Æîü ŸÖ²Ö ˆÃÖ ×ãÖ×ŸÖ ´Öë ÛãúÞ›ü»Öß 
¯Ö¸ü »ÖÝÖÖ ²Ö»Ö-†Ö‘ÖæÞÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … ×ÛúÃÖ ×¾Ö®µÖÖÃÖ ´Öë µÖÆü ÛãúÞ›ü»Öß Ã£ÖÖµÖß ÃÖÖ´µÖÖ¾ÖãÖÖ ´Öë ÆüÖêÝÖß ?
A square loop of side 20 cm carrying current of 1A is kept near an infinite long 
straight wire carrying a current of 2A in the same plane as shown in the figure. 
Calculate the magnitude and direction of the net force exerted on the loop due to the 
current carrying conductor. 
OR 
A square shaped plane coil of area 100 cm
2
 of 200 turns carries a steady current of 5A. 
It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of 
the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the 
direction of the field. In which orientation will the coil be in stable equilibrium ? 
9. ˆ®Ö ¾ÖîªãŸÖ “Öã´²ÖÛúßµÖ ×¾Ö×Ûú¸üÞÖÖë ÛúÖ ®ÖÖ´Ö ×»Ö×ÜÖ‹ (i) ו֮ÖÛúÖ ˆ¯ÖµÖÖêÝÖ ÛïúÃÖ¸ü Ûúß ÛúÖê׿ÖÛúÖ†Öë ÛúÖê ®Ö™ü Ûú¸ü®Öê ´Öë
×ÛúµÖÖ •ÖÖŸÖÖ Æîü, (ii) ו֮ÖÃÖê Ÿ¾Ö“ÖÖ ŸÖÖ´ÖÏ ¸ÓüÝÖ Ûúß ÆüÖê •ÖÖŸÖß Æîü, (iii) ¯Ö飾Öß Ûúß ˆÂÞÖŸÖÖ ²Ö®ÖÖ‹ ¸üÜÖŸÖê Æïü … 2
‡®Ö´Öë ÃÖê ×ÛúÃÖß ‹Ûú ¯ÖÏÛúÖ¸ü Ûúß ŸÖ¸ÓüÝÖÖë ÛúÖê ˆŸ¯Ö®®Ö Ûú¸ü®Öê Ûúß ×¾Ö×¬Ö ÛúÖ ÃÖÓõÖê¯Ö ´Öë ¾ÖÞÖÔ®Ö Ûúßו֋ …
Name the types of e.m. radiations which (i) are used in destroying cancer cells,
(ii) cause tanning of the skin and (iii) maintain the earth’s warmth.
Write briefly a method of producing any one of these waves.
55/1 5 [P.T.O. 
10. ×“Ö¡Ö ´Öë ÆüÖ‡›ÒüÖê•Ö®Ö ¯Ö¸ü´ÖÖÞÖã ÛúÖ ‰ú•ÖÖÔ ÃŸÖ¸ü †Ö¸êüÜÖ ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü : 2
(a) ¾ÖÆü ÃÖÓÛÎú´ÖÞÖ –ÖÖŸÖ Ûúßו֋ וÖÃÖ´Öë 496 nm ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ±úÖê™üÖò®Ö ÛúÖ ˆŸÃÖ•ÖÔ®Ö ÆüÖêŸÖÖ Æîü …
n = 4
n = 3
n = 2
n = 1 
(b) ×ÛúÃÖ ÃÖÓÛÎú´ÖÞÖ Ûêú ÃÖÓÝÖŸÖ †×¬ÖÛúŸÖ´Ö ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ×¾Ö×Ûú¸üÞÖ ˆŸÃÖÙ•ÖŸÖ ÆüÖêŸÖê Æïü ? †¯Ö®Öê ˆ¢Ö¸ü Ûúß ¯Öã×™ü 
Ûúßו֋ … 
The figure shows energy level diagram of hydrogen atom.  
(a)  Find out the transition which results in the emission of a photon of wavelength 
496 nm. 
n = 4
n = 3
n = 2
n = 1 
(b) Which transition corresponds to the emission of radiation of maximum 
wavelength ? Justify your answer. 
ÜÖÞ›ü – ÃÖ
Section – C 
11. (a) ×ÛúÃÖß “ÖÖ»ÖÛú ´Öë ¯ÖϾÖÖ×ÆüŸÖ ¬ÖÖ¸üÖ ‘I’ †Öî¸ü ‡»ÖꌙÒüÖò®Ö Ûêú †¯Ö¾ÖÖÆü ¾ÖêÝÖ 
?
?
d 
Ûêú ²Öß“Ö ÃÖÓ²ÖÓ¬Ö ¾µÖ㟯֮®Ö Ûúßו֋ … 3 
(b) ×“Ö¡Ö ´Öë ×ÛúÃÖß ŸÖÖ¸ü Ûúß †®Öã¯ÖÏÃ£Ö ÛúÖ™ü ÃÖê ¯ÖϾÖÖ×ÆüŸÖ ¬ÖÖ¸üÖ ‘I’ †Öî¸ü ÃÖ´ÖµÖ ‘t’ Ûêú ²Öß“Ö ÝÖÏÖ±ú ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü … 
‡ÃÖ ÝÖÏÖ±ú ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ŸÖÖ¸ü ÃÖê 10s ´Öë ¯ÖϾÖÖ×ÆüŸÖ †Ö¾Öê¿Ö –ÖÖŸÖ Ûúßו֋ …
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